Computing $ \lim_{n\rightarrow \infty}\left(\frac{1^n+2^n+\cdots \cdots +n^n}{n^n}\right)$

Question

Computing $\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1^n+2^n+\cdots \cdots +n^n}{n^n}\right)$

Attempt: $\displaystyle \lim_{n\rightarrow \infty}\bigg[\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots \cdots +\left(\frac{n-1}{n}\right)^n+\left(\frac{n}{n}\right)^n\bigg] = 1$

Is my answer right? If not, then could someone help me? Thanks.

Answer 1

Using $$ \lim_{n\rightarrow \infty} \left(1-\frac{a}{n}\right)^n= e^{-a} $$ Term by term $$ \displaystyle \lim_{n\rightarrow \infty}\bigg[\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots \cdots +\left(\frac{n-1}{n}\right)^n+\left(\frac{n}{n}\right)^n\bigg] = (e^{-n+1}+e^{-n+2}+\cdots +e^{-1}+1) = \frac{1}{1-e^{-1}} = \frac{e}{e-1} \sim 1.58198 $$

Numerically

• $n = 10$, the sum equals 1.49143
• $n = 100$, the sum equals 1.57211.
• $n = 200$, the sum equals 1.57702.

Edit I have to warn that the term by term step is problematic (thanks a lot to the commentators!), as it uses $$ \lim_{n\rightarrow \infty} \frac{1}{n^n} = \frac{e}{e^{n}} $$ I couldn't figure out a good way of explaining this and make the whole process more rigorous (I am from physical sciences instead of mathematics), but I decide to leave the answer here as it should still be partially useful. I will attempt to refine it in the future.